Brieskorn resolutions via algebraic spaces
I’d like to discuss simultaneous resolutions of surfaces from a moduli-theoretic perspective, following Michael Artin’s paper on Brieskorn resolutions.
Artin’s approach to moduli begins with the most desirable aspect of a moduli space, its universal property. That is to say, define a functor the space should represent and then check some local criteria to establish it’s algebraicity. Among the first successes of this method was the Hilbert, Quot, and Picard schemes (see his Formal Moduli I paper) but later Artin would go on to show that modifications (e.g. contractions) could also be represented by algebraic spaces, even if they weren’t representable by schemes. Using this idea, he showed that the category of algebraic spaces algebraically reproduces the category of Moishezon spaces in complex-analytic geometry (see his Formal Moduli II paper).
Here the primary goal will be to prove another algebraic avatar of a result in complex analytic geometry:
Theorem 1 (Brieskorn) Let denote a flat family of algebraic surfaces over
having only rational double points as singularities, then there is an analytic space and a finite surjective morphism
such that
admits a simultaneous resolution of singularities.
(*) To fix ideas, from here onwards let denote a flat, separated and finite-type morphism of schemes over a field whose singular locus
is finite over
and whose geometric fibers are all normal and purely two-dimensional.
Definition 2 Let be as above, a proper morphism
is a resolution if
is smooth and on each fiber the induced morphism
is a minimal resolution of singularities. Two resolutions of
are isomorphic if they are isomorphic over
.
Artin begins by defining a functor on the category of -schemes:
set of resolutions of
upto isomorphism
If we want to understand the extent to which can be simultaneously resolved, we had better understand this functor. The goal is to represent
by an algebro-geometric object since, by definition, its points correspond to resolutions of the relative surface
. In particular, it has a desirable universal property: the surface
can be simultaneously resolved over an
-scheme
if and only if
admits a
-point. In fact, the representability follows from:
Theorem 3 (Artin) Let be as in *, then
is a locally quasi-separated algebraic space.
We leave the proof of this result to a future post and grant it for now. Here is the algebraic analogue of Brieskorn’s result:
Theorem 4 (Artin) Suppose is as in (*) and is a versal deformation of a single rational singularity on
. Let
be the structure morphism and suppose
gets mapped to
. Then the local ring
is nonsingular and the canonical morphism
maps the domain onto an irreducible component of . In particular, if the latter is irreducible, then the morphism is finite and surjective.
Remark: Note that in order for the last sentence of the Theorem to apply, we need the Henselian local rings of the base scheme to be unibranch, a condition which is guaranteed if
is normal. However, by hypothesis if
denotes the category of Artin local rings with residue field
for some
, the map
induced by
, is formally smooth (this is just a reformation of versality). Thus, if
is formally smooth over
(i.e. if the deformations of
are unobstructed, a situation that holds if
has rational double or triple points) we may deduce that
is unibranch at
and hence that the associated maps on local rings are both finite and surjective, a direct analogue of Brieskorn’s analytic result.
In the next post we’ll explain the proof of Theorem 4. Next, we’ll give some elementary observations and then conclude with two interesting examples.
Proposition 5: Let be as in (*) above and suppose
is the resolution functor with
mapping to
.
(a) If is a morphism, then there is a natural identification
.
(b) The structure morphism induces a bijection on
-valued points for any field
.
(c) If is smooth, then the functor
is the singleton functor i.e. the structure morphism
is an isomorphism.
(d) If is smooth at a point
, then the morphism
is an isomorphism in an open neighborhood of
.
(e) The induced map is finite for any
.
(f) If is a finite open cover of
such that
is smooth for all
, then the natural morphism defined by restriction
is an isomorphism of algebraic spaces.
Proof: Both (a) and (c) are easy to see so we omit their proofs. For (b) recall that minimal resolutions of surfaces always exist and are unique upto unique isomorphism. As for (d), note that by the finiteness of , there is an open neighborhood of
over which
is smooth, combining this with (a) yields the result:
.
For (e), note that we may replace with
since the formation of
is compatible with restriction, so we assume
is the spectrum of a Henselian local ring. Now, let
Lastly, for (f) we need to show that the natural morphism of functors
is an equivalence. Indeed, if two resolutions and
become isomorphic over the open cover
, say, via isomorphisms
, then these morphisms patch because the
are all isomorphisms over
. This shows injectivity, surjectivity follows because we may patch together a collection of resolutions
. Indeed, for the glueing datum we use the isomorphisms
(which easily satisfies the cocycle condition). Therefore the resulting resolution of
restricts to the point defined by the resolutions
. QED
The extent to which a given relative surface can be resolved simultaneously is reflected in the geometry of the morphism
. For example, the worst case scenario is when
is the disjoint union of the points in
:
Example: A necessary condition for the simultaneous resolution of a family of surfaces is that each of their minimal resolutions must have the same arithmetic genus. In particular, a degeneration of a smooth family of surfaces has a chance of being simultaneously resolved only if the singularities on the special fiber are rational. Consider for instance, a pencil consisting of smooth surfaces for all and which yields the cone over an elliptic curve at
. The minimal resolution has an elliptic curve as its exceptional locus which means the arithmetic genus of the surface must change along the resolution. Since the original family is flat it means the smooth members of the family and the resolution have differing Euler characteristics and hence they cannot appear in a flat family together. Thus, a simultaneous resolution cannot exist in this case and so if we replace the base of the pencil with the spectrum of a DVR specializing to
, we see that
must be a disjoint union of points.
Example: It is worth noting that the resulting algebraic space need not be a scheme, even in simple cases. Let be the spectrum of a DVR and
a degeneration of a smooth quartic surface in
whose general fiber
has
and whose special fiber
has a single node
. We claim that
cannot be simultaneously resolved in the category of schemes. If so, this also shows that
cannot be represented by a scheme since the universal resolution
is a projective morphism. Indeed, the dual of the line bundle associated to the exceptional curve is a relatively ample line bundle. Thus, if
was a scheme,
and hence
would be as well, a contradiction.
Let be a resolution so that
is an isomorphism and
is minimal. Next, choose a hyperplane section
so that it’s closure
misses
and is therefore isomorphic to its preimage in
. It follows that
. If
was a scheme there would be an open affine subset
meeting
, the complement would also be a Cartier divisor
which is flat over
. Indeed, we know any generic point
of
has height one and so if
, it must be the generic point of the
because the dimension of its closure must be two. This would mean that
contains the special fiber, a contradiction by our choice of
and so
is
-flat. We now have
for some positive integer
. To see this, observe that
is trivial over the generic fiber (for some positive integer
) so we get the desired isomorphism by upper-semicontinuity. But the linear equivalence of
and
implies
or, in other words, that
and
are disjoint. This means the projective scheme
must live entirely inside the open affine subset
, a contradiction.
That’s it for now, in the next post I will explain the proof of Theorem 4.